Question

If log[x+y/3]=1/2[logx+logy]=1/2[logx+logy] find the value of x/y+y/x

Answer 1

Question:

if $\sf\:\log( \frac{x + y}{3} ) = \frac{1}{2} ( \log(x) + \log_{}(y) )$

find the value of $\sf\:\dfrac{ x}{y} + \dfrac{ y}{x}$

Solution :

$\sf\:\log( \frac{x + y}{3} ) = \frac{1}{2} ( \log(x) + \log_{}(y) )$

use property :

log(a) + log (b) = log (ab)

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$\sf\:\implies\:log( \frac{x + y}{3} ) = \frac{1}{2} log(xy)$

$\sf\:\implies\:log( \frac{x + y}{3} ) - \frac{1}{2} log(xy) = 0$

use property log (a/b) = log a- log b

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$\sf\:\implies\:log(x + y) - log(3) - \frac{1}{2} log(xy) = 0$

$\sf\:\implies\:2 log(x + y) - log(xy) = 2 log(3)$

use property :

$\sf\:\log(a) {}^{n} = n log(a)$

$\sf\:\implies\:\log(x + y) {}^{2} - log(xy) = log(3) {}^{2}$

Now use property ;

log (x/y ) = logx -logy

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$\sf\:\implies\:\log( \dfrac{(x + y) {}^{2} }{xy} ) = log(9)$

$\sf\:\implies\:\dfrac{(x + y ){}^{2} }{xy} = 9$

$\sf\:\implies\dfrac{x {}^{2} + y {}^{2} + 2xy}{xy} = 9$

$\sf\:\implies\dfrac{x {}^{2} }{xy} + \dfrac{y {}^{2} }{xy} + \dfrac{2xy}{xy} = 9$

$\bf\:\implies\dfrac{x}{y} + \dfrac{y}{x} = 7$

it is the required solution!

Properties of logarithm :

$\sf1) log(a) + log(b) = log(b)$

$\sf2) log( \frac{a}{b} ) = log(a) - log(b)$

$\sf3) log(a) {}^{n} = n log(a)$

$\sf4) log_{x}(x) = 1$

Answer 2

log(x+y/3) = 1/2(log x + log y)

log(x+y/3) = 1/2 log (xy)

log(x+y/3) = log (xy)1/2

log(x+y/3) = log (√xy)

REMOVE LOG ON BOTH SIDES

(x+y/3) = √xy

SQUARING ON BOTH SIDES

(x+y/3)² = (√xy)²

(x+y)²/(3)² = xy

(x+y)²/9 = xy

(x+y)² = 9xy

x²+y²+2xy = 9xy

x²+y² = 9xy-2xy

x²+y² = 7xy

x²+y²/xy = 7

x²/xy + y²/xy = 7

x/y + y/x = 7

HOPE THIS HELPS U MATE...